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Chapter 8: Problem 15

Solve the following initial value problems. $$y^{\prime}(t)-3 y=12, y(1)=4$$

Short Answer

Expert verified
Question: Solve the initial value problem $$y^{\prime}(t)-3 y=12$$ with the initial condition $$y(1)=4$$. Answer: The particular solution to the given initial value problem is: $$y(t) = -\frac{4}{3} +\frac{16}{3e^3}e^{3t}$$

Step by step solution

01

Identify the differential equation

The given differential equation is a first-order linear ordinary differential equation, and it is written in the standard form: $$y^{\prime}(t) - 3y = 12$$
02

Determining the integrating factor

To find the integrating factor, we first identify the function P(t) in the standard form of the given differential equation: $$y^{\prime}(t) + P(t)y = Q(t)$$ In our case, P(t)=-3 and Q(t)=12. Now, we compute the integrating factor, which is given by the formula: Integrating factor = $$e^{\int -3 dt}$$ Now, let's compute the integral: $$\int -3 dt = -3t + c$$ We don't need the constant "c" in this case because it won't affect our integrating factor. Therefore: Integrating factor = $$e^{-3t}$$
03

Multiply both sides of the differential equation with the integrating factor

We multiply the given differential equation and the integrating factor: $$e^{-3t}(y^{\prime}(t) - 3y) = 12e^{-3t}$$
04

Solve the resulting differential equation

The left side of the new differential equation becomes the derivative of the product of integrating factor and y(t): $$\frac{d}{dt}(ye^{-3t})=12e^{-3t}$$ Now we integrate both sides with respect to t: $$\int\frac{d}{dt}(ye^{-3t})dt = \int 12e^{-3t} dt$$ On the left side, the integral of a derivative is simply the original function, so we get: $$ye^{-3t} = \int 12 e^{-3t}dt + C$$ Now we integrate the right side: $$\int 12e^{-3t} dt = -\frac{4e^{-3t} + C}{3}$$ So the equation becomes: $$ye^{-3t} = -\frac{4e^{-3t}}{3} + C$$ Now, we multiply both sides by $$e^{3t}$$ to isolate y(t): $$y(t) = -\frac{4}{3} + Ce^{3t}$$
05

Use the initial condition to find C

We know that y(1) = 4, so we plug these values into the equation: $$4 = -\frac{4}{3} + Ce^{3(1)}$$ Solve for C: $$C = \frac{4 + \frac{4}{3}}{e^3}$$ $$C =\frac{16}{3e^3}$$
06

Write down the particular solution and check the initial condition

Now plug C back into our general solution to find the particular solution: $$y(t) = -\frac{4}{3} +\frac{16}{3e^3}e^{3t}$$ Finally, check the initial condition y(1) = 4: $$y(1) = -\frac{4}{3} +\frac{16}{3e^3}e^{3(1)}$$ $$y(1) = -\frac{4}{3} + \frac{16}{3} = 4$$ The initial condition is satisfied, so the particular solution is correct. The final answer is: $$y(t) = -\frac{4}{3} +\frac{16}{3e^3}e^{3t}$$

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Most popular questions from this chapter

RC circuit equation Suppose a battery with voltage \(V\) is connected in series to a capacitor (a charge storage device) with capacitance \(C\) and a resistor with resistance \(R\). As the charge \(Q\) in the capacitor increases, the current \(I\) across the capacitor decreases according to the following initial value problems. Solve each initial value problem and interpret the solution. a. \(I^{\prime}(t)+\frac{1}{R C} I(t)=0, I(0)=\frac{V}{R}\) b. \(Q^{\prime}(t)+\frac{1}{R C} Q(t)=\frac{V}{R}, Q(0)=0\)

Consider a loan repayment plan described by the initial value problem $$B^{\prime}(t)=0.03 B-600, \quad B(0)=40,000$$ where the amount borrowed is \(B(0)=\$ 40,000,\) the monthly payments are \(\$ 600,\) and \(B(t)\) is the unpaid balance in the loan. a. Find the solution of the initial value problem and explain why \(B\) is an increasing function. b. What is the most that you can borrow under the terms of this loan without going further into debt each month? c. Now consider the more general loan repayment plan described by the initial value problem $$B^{\prime}(t)=r B-m, \quad B(0)=B_{0}$$ where \(r>0\) reflects the interest rate, \(m>0\) is the monthly payment, and \(B_{0}>0\) is the amount borrowed. In terms of \(m\) and \(r,\) what is the maximum amount \(B_{0}\) that can be borrowed without going further into debt each month?

Consider the following pairs of differential equations that model a predator- prey system with populations \(x\) and \(y .\) In each case, carry out the following steps. a. Identify which equation corresponds to the predator and which corresponds to the prey. b. Find the lines along which \(x^{\prime}(t)=0 .\) Find the lines along which \(y^{\prime}(t)=0\) c. Find the equilibrium points for the system. d. Identify the four regions in the first quadrant of the xy-plane in which \(x^{\prime}\) and \(y^{\prime}\) are positive or negative. e. Sketch a representative solution curve in the xy-plane and indicate the direction in which the solution evolves. $$x^{\prime}(t)=-3 x+x y, y^{\prime}(t)=2 y-x y$$

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of \(t.\) Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t.\) Use this idea to solve the following initial value problems. $$t y^{\prime}(t)+y=1+t, y(1)=4$$

Consider the general first-order linear equation \(y^{\prime}(t)+a(t) y(t)=f(t) .\) This equation can be solved, in principle, by defining the integrating factor \(p(t)=\exp \left(\int a(t) d t\right) .\) Here is how the integrating factor works. Multiply both sides of the equation by \(p\) (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes $$p(t)\left(y^{\prime}(t)+a(t) y(t)\right)=\frac{d}{d t}(p(t) y(t))=p(t) f(t).$$ Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor. $$y^{\prime}(t)+\frac{2 t}{t^{2}+1} y(t)=1+3 t^{2}, \quad y(1)=4$$
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